Problem: $\overline{AC}$ is $15$ units long $\overline{BC}$ is $8$ units long $\overline{AB}$ is $17$ units long What is $\sec(\angle ABC)?$ $A$ $C$ $B$ $15$ $8$ $17$
Answer: $\sec(\angle ABC) = \dfrac{1}{\cos(\angle ABC)}$ How can we find $\cos(\angle ABC)$ SOH CAH TOA osine = djacent over ypotenuse Adjacent $= \overline{BC} = 8$ Hypotenuse $= \overline{AB} = 17$ $\cos(\angle ABC) = \dfrac{8}{17}$ $\sec(\angle ABC) = \dfrac{1}{\cos(\angle ABC)} = \dfrac{17}{8}$